Question

I hava a class

class vlarray {
public:
    double *p;
    int size;

    vlarray(int n) {
        p = new double[n];
        size = n;
        for(int i = 0; i < n; i++)
            p[i] = 0.01*i;
    }

    ~vlarray() {
        cout << "destruction" << endl;
        delete [] p;
        size = 0;
    }
};

when I use in main

int main() {
    vlarray a(3);
    {
        vlarray b(3);
        b.p[0] = 10;
        for(int i = 0; i < 3; i++) {
            cout << *(b.p+i) << endl;
        }
        a = b;
    }    // the magic happens here deallocation of b
    for(int i = 0; i < 3; i++) {
        cout << *(a.p+i) << endl;
    return 0;
}

when b deallocated smth happens to a .. what is the problem, why that problem occurs and how to avoid such type of problems?

Answer 1

_{There seems to be some confusion amongst the existing answers on this question, so I am going to jump in.}

---

Immediate issue

Your primary issue is that you have not defined your own copy assignment operator. Instead, the compiler generates a naive one for you, so that when you run a = b, the pointer inside b is copied into a. Then, when b dies and its destructor runs, the pointer now inside a is no longer valid. In addition, a‘s original pointer has been leaked.

Your own copy assignment operator will need to delete the existing array, allocate a new one and copy over the contents from the object you’re copying..

More generally

Going further, you will also need to define a few other things. This requirement is neatly summed-up as the Rule of Three (C++03) or Rule of Five (C++11), and there are plenty of explanations online and in your favourite, peer-recommended C++ book that will teach you how to go about satisfying it.

Or, instead…

Better still, you could start using an std::vector instead of manually allocating everything, and avoid this entire mess:

struct vlarray {
    std::vector<double> p;

    vlarray(int n) {
        p.resize(n);
        for(int i = 0; i < n; i++)
            p[i] = 0.01*i;
    }
};